Flux — Question 10

Section 23-1 · Physics pages 3650–3651

Question 10 Three independent flux measurements reconstruct a uniform flow Three small planar windows sample the same uniform velocity. Each has area .100 m2. Their normals are ˆx, ˆy and (ˆx + ˆy + ˆz)/ √ 3. Signed fluxes are .200, −.100 and .300/ √ 3 m3/s, respectively. ˆn1 ˆn2 ˆn3 Schematic of three independent measurement orientations; the third has a nonzero z component. (a) Reconstruct the three Cartesian velocity components and its speed. (b) Predict the flux through a fourth equal-area window with normal (ˆx −ˆz)/ √ 2. Explain whether its zero reading would imply zero speed.
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Question 10 - Solution Three independent flux measurements reconstruct a uniform flow ˆn1 ˆn2 ˆn3 (a) Method and calculation vx = Q1/A = 2.00, vy = Q2/A = −1.00 (m/s). vx + vy + vz = √ 3Q3/A = 3.00 ⇒vz = 2.00 m/s. (b) Interpretation and further calculation v = √ 22 + (−1)2 + 22 = 3.00 m/s. Q4 = A √ 2(vx −vz) = 0. The fourth normal is perpendicular to the reconstructed velocity; its zero reading is compatible with a moving fluid. Physical check Reconstruction requires independent normals. Repeating parallel measurements cannot recover missing transverse components.

Question 10

Three independent flux measurements reconstruct a uniform flow

Three small planar windows sample the same uniform velocity. Each has area .100.100 m2^2. Their normals are 𝒙̂\hat{\mathbf x}, 𝒚̂\hat{\mathbf y} and (𝒙̂+𝒚̂+𝒛̂)/3(\hat{\mathbf x}+\hat{\mathbf y}+\hat{\mathbf z})/\sqrt3. Signed fluxes are .200.200, −.100-.100 and .300/3.300/\sqrt3 m3^3/s, respectively.

Diagram shown in the original SVG worksheet.

  1. Reconstruct the three Cartesian velocity components and its speed.

  2. Predict the flux through a fourth equal-area window with normal (𝒙̂−𝒛̂)/2(\hat{\mathbf x}-\hat{\mathbf z})/\sqrt2. Explain whether its zero reading would imply zero speed.

Question 10 - Solution

Three independent flux measurements reconstruct a uniform flow

Diagram shown in the original SVG worksheet.

(a) Method and calculation

vx=Q1/A=2.00,vy=Q2/A=−1.00(m/s).v_x=Q_1/A=2.00,\quad v_y=Q_2/A=-1.00\qquad(\mathrm{m/s}).vx+vy+vz=3Q3/A=3.00⇒vz=2.00m/s.v_x+v_y+v_z=\sqrt3Q_3/A=3.00\Rightarrow v_z=2.00\,\mathrm{m/s}.

(b) Interpretation and further calculation

v=22+(−1)2+22=3.00m/s.v=\sqrt{2^2+(-1)^2+2^2}=3.00\,\mathrm{m/s}.Q4=A2(vx−vz)=0.Q_4=\frac A{\sqrt2}(v_x-v_z)=0.The fourth normal is perpendicular to the reconstructed velocity; its zero reading is compatible with a moving fluid.

Physical check

Reconstruction requires independent normals. Repeating parallel measurements cannot recover missing transverse components.